Question

How would be the three register each 9 ohms show the equivalent resistance will be 13.5 and 6 ohms

Answer 1

1. equivalent resistance = 13.5 Ω

Join two resistors in parallel and one in series as shown in 1st figure

We know, in parallel ,

1/R' = 1/R₁ + 1/R₂

Here , R₁ = R₂ = R₃ = 9Ω

Now, 1/R' = 1/9Ω + 1/9Ω ⇒ R' = 9/2Ω = 4.5 Ω

So, equivalent resistance , Req = R' + R₃ [ ∵ R' and R₃ in series ]

∴ Req = 4.5Ω + 9Ω = 13.5Ω

2. Equivalent resistance = 6Ω

two resistors are in series which is parallel with 3rd resistor as shown in 2nd figure .

now, R' = R₁ + R₂ = 9Ω + 9Ω = 18Ω [ R₁ and R₂ are in series ]

now, equivalent resistance, Req = R'R₃/(R' + R₃) [ R' and R₃ are in parallel ]

∴ Req = 18 × 9/(18 + 9) = 18 × 9/27 = 6

Answer 2

Answer:

We can connect the three resistors in the following way:

(a) First two resistance connected in parallel and the resulted is connected with another resistor in series. Equivalent resistance of combination is 13.5 ohm

(b) First two resistance connected in series and the resulted is connected with another resistor in parallel. Equivalent resistance of combination is 6 ohm.

We know, for 2 resistances, R₁ and R₂, the equivalent resistance:

(1) When connected in series is

R =  R₁ + R₂

(2) When connected in parallel is

R = (R₁ × R₂)/(R₁ + R₂)

Here, in the question, R₁ = R₂ = R₃ = 9Ω

(a) When R₁ and R₂ are first connected in parallel,

Equivalent resistance R' = (9 × 9)/(9 + 9)Ω = 81/18Ω = 4.5Ω

Again when the third resistance R₃ is connected to R' in series, we get equivalent resistance as R = (4.5 + 9)Ω = 13.5Ω

(b) When R₁ and R₂ are first connected in series,

Equivalent resistance R' = (9 + 9)Ω = 18Ω

Again when the third resistance R₃ is connected to R' in parallel, we get equivalent resistance as R = (18 × 9)/(18 + 9)Ω = 6Ω

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